In the figure shown `V_(1),V_(2),V_(3)` are `AC` voltmeters and `A` is `AC` ammeter. The readings of `V_(1), V_(2), V_(3)` and `10 V, 20 V, 20 V, 2A` respectively. If the inductor is short circuited, then

Options

A. the reading of `V_(1)` is `2 sqrt(5) V`
B. the reading of `V_(2)` is `4 sqrt(5) V`
C. the reading of `V_(2)` is `2 sqrt(5) V`
D. the value of `A` is `(2)/(sqrt(5)) A`

Answer

A,B,D

Explanation

(a) Impedance of circuit `= sqrt(R^(2) + (X_(C ) + X_(L))^(2))`
`Z = sqrt(8^(2) + (8 - 2)^(2)) = 10 Omega`
(b) The current leads in phase by `( :.' X_(c ) gt X_(L)) phi = 37^(@)`
`:. i = (10 cos (100 pi t + 37^(@))/(Z) = cos (100 pi t + 37^(@))`
The instantaneous potential difference across `AB` is
`= I_(m) = (X_(C ) - X_(L)) cos (100 pi t + 37^(@) - 90^(@))`
`= 6 cos (100 pi t - 53^(@))`
The instantaneous petential difference across `AB` is half of source voltage.
`= 6 cos (100 pi t - 53^(@)) = 5 cos 100 pi t`
Solving, `Tan (100 pi t) = (17)/(24)`
`:. t = (1)/(100 pi) Tan^(-1) ((17)/(24))`